Assume the following summation,
$$ \sum_{i=0}^{1000}\left(-1\right)^{i}{1000 \choose i}\left(100 - i\right)^{500}. $$
I know that this summation is zero, $0$ ( I've checked it with Maple, though ). But I cannot find any proof for that!. Can you provide any help ?.
P.S. This is not a homework problem, just a question one of my friend asked me.
If $(Δ p)(x)=p(x+1)-p(x)$ denotes the step-one difference operator, your expression is equal to $$0=(Δ^{1000} p)(x)\quad\text{where}\quad p(x)=(100-x)^{500}$$
Since each application of $Δ$ lowers the degree of the polynomial by one, already $(Δ^{501} p)(x)=0$, and moreso the higher order differences.
If $p(x)=x^k$, then $$ (Δ p)(x)=p(x+1)-p(x)=(x+1)^k-x^k=(x+1)^{k-1}+(x+1)^{k-2}x+...+(x+1)x^{k-2}+x^{k-1} $$ is a polynomial of one degree less, this extends linearly to all polynomials. Thus $\deg(Δ^m p)=\deg p-m$ as long as $m\le\deg p$, and $Δ^{1+\deg p} p=0$.
For the higher order differences on gets the formula $$ (Δ^m p)(x)=\sum_{i=0}^m(-1)^{m-i}\binom{m}{i} p(x+i) $$ This is obviously true for $m=0$, and easy to check for $m=1$, let's do the induction step \begin{align} (Δ^{m+1} p)(x)&=(Δ^m p)(x+1)-(Δ^m p)(x) \\ &=\sum_{i=0}^m(-1)^{m-i}\binom{m}{i} p(x+i+1)-\sum_{i=0}^m(-1)^{m-i}\binom{m}{i} p(x+i) \\ &=\sum_{i=1}^{m+1}(-1)^{m-i+1}\binom{m}{i-1} p(x+i)+\sum_{i=0}^m(-1)^{m-i+1}\binom{m}{i} p(x+i) \\ &=\sum_{i=0}^{m+1}(-1)^{m+1-i}\binom{m+1}{i} p(x+i) \end{align}
Even easier is the derivation using the translation operator $(Tp)(x)=p(x+1)$. Then $Δ=(T-I)$ and $$ Δ^m=(T-I)^m=\sum_{i=0}^m\binom{m}{i}(-1)^{m-i}T^{i}. $$
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{i=0}^{1000}\pars{-1}^{i}{1000 \choose i}\pars{100 - i}^{500} = 0:\ {\large ?}}$
\begin{align} &\sum_{i=0}^{1000}\pars{-1}^{i}{1000 \choose i}\pars{100 - i}^{500} =\sum_{i=0}^{1000}\pars{-1}^{i}{1000 \choose i}\sum_{\ell = 0}^{500} {500 \choose \ell}100^{500 - \ell}\pars{-i}^{\ell} \\[3mm]&=\sum_{\ell = 0}^{500}{500 \choose \ell}\pars{-1}^{\ell}100^{500 - \ell} \underbrace{\sum_{i=0}^{1000}{1000 \choose i}\pars{-1}^{i}i^{\ell}} _{\ds{=\ 0;\quad\mbox{See bellow}}} \end{align}
\begin{align} \pars{1 - z}^{1000}&=\sum_{i = 0}^{1000}{1000 \choose i}\pars{-1}^{i}z^{i} \\ -1000z\pars{1 - z}^{999}&=\sum_{i = 0}^{1000}{1000 \choose i}\pars{-1}^{i}iz^{i} \\ -1000z\pars{1 - z}^{999} + 1000\times999 z^{2}\pars{1 - z}^{998}& =\sum_{i = 0}^{1000}{1000 \choose i}\pars{-1}^{i}i^{2}z^{i}\,, \end{align} If we continue this procedure we get increasing powers of $i$ ( up to $i^{\ell}$ with $\ell \leq 500$ ) in the right member. Then, set $z = 1$ in both members and we'see that all the sums vanish out.